Which point of the sphere $x^2+y^2+z^2=19$ maximize $2x+3y+5z$?

Question

Which point of the sphere $x^2+y^2+z^2=19$ maximize $2x+3y+5z$?

So I assume that there is a point maximizing $2x+3y+5z$. How can I calculate the exact value of this point?

Answer 1

By the Cauchy-Schwarz inequality

$$ 2x+3y+5z \leq \sqrt{x^2+y^2+z^2} \sqrt{2^2+3^2+5^2} = 19 \sqrt{38} $$ and equality is achieved if $(x,y,z)=\lambda(2,3,5)$, i.e. for $\lambda=\frac{1}{\sqrt{2}}$.

Answer 2

You are on a sphere of radius $\sqrt{19}$ centered at the origin. Spherical coordinates is an option: with $x=\sqrt{19}\sin \phi \cos \theta$, $y=\sqrt{19}\sin \phi \sin \theta$ and $z=\sqrt{19}\cos \phi$, the problem boils down to solving $$ \max \quad 2\sqrt{19}\sin \phi \cos \theta+3\sqrt{19}\sin \phi \sin \theta+5\sqrt{19}\cos \phi $$ with $(\theta,\phi)\in [0,2\pi] \times [0,\pi]$.

Answer 3

You should look for a vector $(x,y,z)$, with a norm equal to $19$, such that its inner-product with $(2,3,5)$ is maximum.

$2x+3y+5z=<(2,3,5),(x,y,z)>=||(2,3,5)||\times||(x,y,z)||\times cos(\theta)=\sqrt{38}\times 19\times cos(\theta)$

Therefore,just focus on $cos(\theta)$ to maximise the function.